MAT 5161 Computational Methods Solution of linear and nonlinear system of equations: . (8 hours) Direct methods: Gauss Jordon method, Crouts (LU decomposition) method, Cholesky Decomposition method and Thomas Algorithm for tridiagonal systems. Indirect Methods(Iterative methods): Gauss Seidal and successive over relaxation, Newton Raphson method (system of non-linear equations), Birgevieta method, Bairstow’s methods. Numerical Solution of Ordinary Differential Equations: (8 hours) Initial Value Problems: Single step methods, Runge-Kutta method, Runge Kutta method for simultaneous differential equations, Runge Kutta method for higher order differential equations, Shooting method. Multi step methods: Adam Bashforth’s predictor corrector method, Milne’s predictor and corrector method. Boundary Value Problems : (10 hours) Finite difference method (IVP) Numerical Solution of Partial Differential Equations- Elliptic P.D.E., Parabolic P.D.E., Hyperbolic P.D.E by explicit finite difference method , ADE and ADI method. Fourier Integrals, Fourier transforms and properties (8 hours) Laplace transforms of elementary functions – inverse Laplace transforms Mathematical Modelling: (8 hours) (10 hours) MM with ordinary/partial/higher order differential equations: Models in arms race, compartment models, heat flow problems and vibration of strings. Modelling through graphs. Solution of linear system of equations: Consider the System of linear equations ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ + ๐๐๐ ๐๐ = ๐๐ ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ + ๐๐๐ ๐๐ = ๐๐ โฎ ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ ๐๐๐ ๐๐ = ๐๐ Gauss Jordon method: ๏ In this method, we are reducing the system to a diagonal matrix Apply Gauss-Jordon method to solve the equations ๐+๐+๐=๐ ๐๐ − ๐๐ + ๐๐ = ๐๐ ๐๐ + ๐๐ + ๐๐ = ๐๐ Solution: We can rewrite the given equations as ๐ ๐ ๐ ๐น๐ → ๐น๐ − ๐๐น๐ ; ๐น๐ → ๐น๐ − ๐๐น๐ ๐ ๐ ๐ ๐ −๐ ๐ ๐ = ๐๐ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐ −๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ = −๐ ๐ ๐ ๐๐ ๐น๐ → ๐น๐ + ๐น๐ → ๐ ๐ ๐ ๐น๐ ๐ ๐น๐ ; −๐ ๐น๐ → ๐ −๐ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๐น๐ ๐๐ ๐ ๐ ๐ ๐ ๐น๐ → ๐น๐ + ๐น๐ ๐น๐ → ๐น๐ − ๐น๐ − ๐น๐ ๐ ๐ ๐ ๐ ๐ ๐ = −๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ = ๐ − ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ = ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ = ๐ ๐ ๐ ๐ ๐ ๐ ๐ −๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ = −๐ ๐ ๐ ๐๐ Apply Gauss-Jordon method to solve the equations ๐ + ๐๐ + ๐๐ = ๐๐ ๐ + ๐๐ + ๐๐ = ๐๐ ๐ + ๐๐ + ๐๐ = ๐๐ LU Decomposition (Factorization method): This method is based on the Fact that a square matrix A can be factorized into the form LU, where L is unit lower triangular matrix and U is upper triangular, if all the principle minors of A are non-singular. i.e., ๐11 ≠ 0, ๐11 ๐21 ๐11 ๐12 ๐21 ≠ 0, ๐22 ๐31 ๐12 ๐13 ๐22 ๐23 ≠ 0, etc ๐32 ๐33 Also, such a factorization if exists, is unique. ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ + ๐๐๐ ๐๐ = ๐๐ ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ + ๐๐๐ ๐๐ = ๐๐ โฎ ๐๐๐ ๐๐ + ๐๐๐ ๐๐ + โฏ ๐๐๐ ๐๐ = ๐๐ Which can be written, ๐จ๐ฟ = ๐ฉ Let ๐จ = ๐ณ๐ผ Eq (1) → ๐ณ๐ผ๐ฟ = ๐ฉ ๐๐๐ _____ (1) where ๐จ = ๐๐๐ โฎ ๐๐๐ ๐๐๐ … ๐๐๐ … ๐๐๐ … ๐๐ ๐๐ ๐๐๐ ๐๐๐ , ๐ฟ = ๐๐ , ๐ฉ = ๐๐ โฎ โฎ ๐๐๐ ๐๐ ๐๐ Take ๐๐ = ๐____ (2) Eq (2) → ๐ณ๐ = ๐ฉ ____ (3) Using (3), find ๐ฆ1 , ๐ฆ2 , … , ๐ฆ๐ by forward substitution. Then using relation (2) ๐๐ = ๐, find ๐ฅ1 , ๐ฅ2 , … , ๐ฅ๐ . Solve the equations ๐๐๐ + ๐๐๐ + ๐๐ = ๐ ๐๐ + ๐๐๐ + ๐๐๐ = ๐ ๐๐๐ + ๐๐ + ๐๐๐ = ๐ by LU decomposition Ans: ๐ณ๐ผ = ๐จ 1 ๐21 ๐31 ๐ข11 0 1 ๐32 0 ๐ข11 0 0 1 0 ๐ข12 ๐ข22 0 ๐ข13 2 ๐ข23 = 1 ๐ข33 3 ๐ข12 ๐ข13 ๐21 ๐ข11 ๐21 ๐ข12 + ๐ข22 ๐21 ๐ข13 + ๐ข23 ๐31 ๐ข11 ๐31 ๐ข12 + ๐32 ๐ข22 ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 3 1 2 3 1 2 2 = 1 3 3 1 2 3 1 2 ๐ข11 ๐ข12 ๐ข13 ๐21 ๐ข11 ๐21 ๐ข12 + ๐ข22 ๐21 ๐ข13 + ๐ข23 ๐31 ๐ข11 ๐31 ๐ข12 + ๐32 ๐ข22 ๐ข11 = 2 ๐ข12 = 3 ๐ข13 = 1 1 (3) + ๐ข22 = 2 2 1 ๐ข22 = 2 ๐31 ๐ข12 + ๐32 ๐ข22 = 1 2 2 3 2 3 + ๐32 3 1 2 3 1 2 ๐21 ๐ข13 + ๐ข23 = 3 ๐21 ๐ข12 + ๐ข22 = 2 ๐21 ๐ข11 = 1 โน ๐21 2 = 1 1 โน ๐21 = ๐31 ๐ข11 = 3 โน ๐31 2 = 3 3 โน ๐31 = ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 2 = 1 3 1 (1) + ๐ข23 = 3 2 5 ๐ข22 = 2 ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 = 2 ๐ข33 = 18 1 9 = 1 โน ๐32 = 1 − 2 = −7 2 2 ๐ฟ๐ = ๐ต 1 0 0 ๐ฆ1 9 1Τ 1 0 ๐ฆ2 = 6 2 3Τ ๐ฆ3 8 2 −7 1 Solving ๐ฆ1 = 9 1 ๐ฆ1 + ๐ฆ2 = 6 2 1 (9) + ๐ฆ2 = 6 2 3 โน ๐ฆ2 = 2 ๐๐ = ๐ 2 3 1 0 2 0 0 1 ๐ฅ 9 5 ๐ฅ1 2 = 3/2 2 ๐ฅ3 5 18 3 ๐ฆ1 − 7๐ฆ2 + ๐ฆ3 = 8 2 โน ๐ฆ3 = 5 ๐๐ ๐ฟ = ๐๐ ๐๐ 5 18๐ฅ3 = 5 โน ๐ฅ3 = 18 1 5 3 29 ๐ฅ2 + ๐ฅ3 = โน ๐ฅ2 = 2 2 2 18 35 2๐ฅ1 + 3๐ฅ2 + ๐ฅ3 = 9 โน ๐ฅ3 = 18 ๐๐เต ๐๐ = ๐๐เต๐๐ ๐เต ๐๐ Crout’s Method: ๐ด๐ = ๐ต → ๐ = ๐ด−1 ๐ต ๐ฟ๐ ๐ = ๐ต To find ๐ด−1 , we first find L and U using LU = A ๐11 Where ๐ฟ = ๐21 ๐31 0 ๐22 ๐32 0 0 ๐33 1 and U= 0 0 ๐ข12 1 0 ๐ข13 ๐ข23 1 After finding L and U, since ๐ฟ๐ฟ−1 = ๐ผ We take ๐ฟ๐ฟ−1 = ๐ผ Similarly, ๐๐ −1 = ๐ผ Now, find ๐จ−๐ = ๐ณ๐ผ ๐11 ๐21 ๐31 ๐22 ๐32 ๐33 1 0 0 ๐ข12 1 0 ๐ข13 ๐ข23 1 −๐ 0 0 0 = ๐ผ−๐ ๐ณ−๐ And hence, we get ๐ฟ = ๐จ−๐ ๐ฉ ๐๐๐ ๐๐ −1 = ๐ผ, ๐ฅ11 ๐ฅ21 ๐ฅ31 0 ๐ฅ22 ๐ฅ32 1 ๐ฆ12 0 1 0 0 0 1 0 = 0 ๐ฅ33 0 ๐ฆ13 1 ๐ฆ23 = 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 Solve by Crout’s method ๐๐๐ − ๐๐๐ + ๐๐๐ = ๐ 2๐๐ + ๐๐๐ + ๐๐๐ = ๐ −๐๐ + ๐๐ − ๐๐ = −๐ ๐ณ๐ผ = ๐จ 1 ๐21 ๐31 0 1 ๐32 ๐ข11 0 0 1 ๐ข11 0 0 ๐ข12 ๐ข22 0 ๐ข13 2 −2 ๐ข23 = 2 3 ๐ข33 −1 1 4 2 −1 ๐ข12 ๐ข13 ๐21 ๐ข11 ๐21 ๐ข12 + ๐ข22 ๐21 ๐ข13 + ๐ข23 ๐31 ๐ข11 ๐31 ๐ข12 + ๐32 ๐ข22 ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 2 = 2 −1 −2 4 3 2 1 −1 ๐ข11 ๐ข12 ๐ข13 ๐21 ๐ข11 ๐21 ๐ข12 + ๐ข22 ๐21 ๐ข13 + ๐ข23 ๐31 ๐ข11 ๐31 ๐ข12 + ๐32 ๐ข22 ๐ข11 = 2 ๐ข12 = −2 ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 ๐ข13 = 4 ๐21 ๐ข12 + ๐ข22 = 3 ๐21 ๐ข11 = 2 โน ๐21 = 1 2 = 2 −1 −2 4 3 2 1 −1 ๐21 ๐ข13 + ๐ข23 = 2 ๐ข23 = −2 ๐ข22 = 5 ๐31 ๐ข11 = −1 โน ๐31 = −1 2 ๐31 ๐ข12 + ๐32 ๐ข22 = 1 โน ๐32 = 0 ๐31 ๐ข13 + ๐32 ๐ข23 + ๐ข33 = −1 ๐ข33 = 1 We take ๐ฟ๐ฟ−1 1 0 0 1 1 0 and ๐ฟ= −1/2 0 1 =๐ผ 2 −2 U= 0 5 0 0 1 0 0 1 1 1 0 ๐ฅ21 −1/2 0 1 ๐ฅ31 0 1 0 = 0 1 0 0 1 ๐ฅ32 1 + ๐ฅ21 = 0 โน ๐ฅ21 = −1 −1/2 + ๐ฅ31 = 0 โน ๐ฅ31 = 1/2 ๐ฟ−1 1 0 = −1 1 1/2 0 0 0 1 ๐ฅ32 = 0 4 −2 1 0 0 1 0 0 1 ๐๐ −1 = ๐ผ ๐ฆ11 = 1/2 2 −2 0 5 0 0 4 ๐ฆ11 −2 0 0 1 ๐ฆ12 ๐ฆ22 0 2๐ฆ12 − 2๐ฆ22 = 0 ๐ฆ33 =1 2๐ฆ12 − 2๐ฆ22 = 0 โน ๐ฆ12 ๐ผ −๐ 0 0 1 0 0 1 2๐ฆ13 − 2๐ฆ23 + 4๐ฆ33 = 0 5๐ฆ23 − 2๐ฆ33 = 0 โน ๐ฆ23 5๐ฆ22 = 1 โน ๐ฆ22 = 1/5 ๐/๐ = ๐ ๐ ๐ฆ13 1 ๐ฆ23 = 0 ๐ฆ33 0 1 = 5 ๐/๐ −๐/๐ ๐/๐ ๐/๐ ๐ ๐ 2 = 5 2๐ฆ13 − 2๐ฆ23 + 4๐ฆ33 = 0 2 2๐ฆ13 − 2 + 4(1) = 0 5 โน ๐๐๐ = − ๐ ๐ ๐ด−1 = ๐ฟ๐ −1 = ๐ −1 ๐ฟ−1 ๐ด−1 0 0 1/2 1/5 −8/5 1 = 0 1/5 2/5 −1 1 0 1/2 0 1 0 0 1 −1/2 1/5 −8/5 0 1/5 2/5 = 1/2 0 1 ๐ = ๐ด−1 ๐ต −1/2 1/5 −8/5 2 0 1/5 2/5 ๐= 5 1/2 0 1 −5 8 ๐ = −1 −4
